[Scilab-users] Cauchy Integral query

[Heinz Nabielek]

Is there a reason that my round-off error is 2.831D-13 ?

Is there a way to get SciLab to print always 2.831e-13 so that I can copy numbers over to EXCEL?

Heinz


--> function y=f(z)
  > 	y = exp((z.^2))./(z-2)
  > endfunction

-->  fz=intl(0, 2*%pi, 2, 1, f,1e-10) // gives round-off error
 fz  =    2.831D-13 + 343.05029i

--> 2*%pi*%i*%e^4
 ans  =    0. + 343.05029i

Scilab Version: 6.1.1.988271013
macOS Catalina Version 10.15.7 


______________-

> On 05.01.2022, at 14:31, sgougeon at free.fr wrote:
> 
> Hello Lester,
> 
> The integrand is y = exp((z^2))/(z-2), not y = exp((z^2)).
> Then, provided that the (undocumented) absolute tolerance is increased wrt the default one,
> we get the expected result:
> 
> --> function y=f(z)
>>  y = exp((z.^2))./(z-2)
>> endfunction
> 
> --> fz=intl(0, 2*%pi, 2, 1, f,1e-10) // gives round-off error
> fz  = 
>   4.199D-13 + 343.05029i
> 
> --> 2*%pi*%i*%e^4
> ans  =
>   0. + 343.05029i
> 
> Regards
> Samuel
> 
>> ----- Mail d'origine -----
>> De: Lester Anderson
>> Envoyé: Wed, 05 Jan 2022 09:46:47 +0100 (CET)
>> 
>> Hello,
>> 
>> I am trying to understand how to work the Cauchy integral inputs and
>> replicate the results of a published example:
>> 
>> .e.g. Compute the integral of e^(z^2) / (z-2) assumes C is closed
>> (anticlockwise) and z=2 is inside C (a simple circle). The solution should
>> be 2*pi*i*f(2) = 2*pi*i*e^4
>> 
>> In Scilab, the solution is defined from the Cauchy Integral (intl):
>> y = intl(a, b, z0, r, f)
>> a and b are real and z complex
>> 
>> function y=f(z)
>>  y = exp((z^2)) // solution uses f(z) =  e^(z^2)
>> endfunction
>> 
>> fz=intl(0, 2*%pi, 2+0*%i, 1, f) // gives round-off error
>> // z position +2(real z), 0(imaginary z)
>> 
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